hdu 1242 Rescue（优先队列）

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5846    Accepted Submission(s): 2168

Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

 

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

Process to the end of the file.

 

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

 

Sample Input

7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........

 

Sample Output

13

      这题要用到优先队列来进行广搜bfs，不然会出现时间长的先到达，这题是我第一次用优先队列，看看哈。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1242

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <queue>
using namespace std;

int xx[4] = {1, -1, 0, 0};
int yy[4] = {0, 0, 1, -1};

int map[205][205];
char a[205][205];
int N, M, x1, y1, x2, y2;
bool flag;

struct node
{
    friend bool operator < (node a, node b) //优先步数少的
    {
        return a.step > b.step;
    }
    int x, y, step;
}n, m;

int main()
{
    int i, j;
    while(scanf("%d %d", &N, &M) != EOF)
    {
        for(i = 0; i < N; i++)
        {
            getchar();
            for(j = 0; j < M; j++)
            {
                scanf("%c", &a[i][j]);
                map[i][j] = 10000000;   //初始化
                if(a[i][j] == 'r')
                    x1 = i, y1 = j;
                if(a[i][j] == 'a')
                    x2 = i, y2 =j;
            }
        }
        flag = false;
        n.x = x1; n.y = y1; n.step = 0;
        priority_queue<node> Q;     //建立优先队列
        Q.push(n);
        while(!Q.empty())
        {
            m = Q.top();
            Q.pop();
            if(m.x == x2 && m.y == y2)  //到达目标
            {
                flag = true;
                break;
            }
            for(i = 0; i < 4; i++)
            {
                n.x = m.x + xx[i];
                n.y = m.y + yy[i];
                if(n.x>=0 && n.x<N && n.y>=0 && n.y<M && a[n.x][n.y]!='#')
                {
                    if(a[n.x][n.y] == 'x') n.step = m.step + 2; //遇到X时间加2
                    else n.step = m.step + 1;   //否则正常加1
                    if(n.step >= map[n.x][n.y]) continue;
                    map[n.x][n.y] = n.step;
                    Q.push(n);
                }
            }
        }
        if(flag) printf("%d\n", m.step);
        else printf("Poor ANGEL has to stay in the prison all his life.\n");
    }

    return 0;
}

 

评论

1 楼 sgeteternal 2011-08-03  

good！